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Why didn't we involve the 4 Ohm resistor while doing the mesh i2 ?

Voltage Spike
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JuanDoe
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  • A mistake? Slip of the pen? – G36 Dec 17 '20 at 20:52
  • I have no idea, I thought maybe it's because that part is an open circuit, so – JuanDoe Dec 17 '20 at 20:56
  • No, what is the answer for the Thevenin voltage? And \$V_\phi\$ is? – G36 Dec 17 '20 at 20:58
  • Vϕ is 20V, Voc is 120V ,i source(when you make it a short circuit)=40 ohm, Rth is 3 ohm, Vth is 60 Volt – JuanDoe Dec 17 '20 at 21:05
  • I meant, what is the "parameter" for \$V_\phi = 4\Omega \times I_1 \times What\$ – G36 Dec 17 '20 at 21:07
  • Im sorry I still don't understand but Vϕ =4(i1-i2) , but i assume you already know this, I didn't get the parameter part though – JuanDoe Dec 17 '20 at 21:11
  • But \$V_\phi\$ is a Voltage controlled Voltage source, so you need to know the "control parameter" value (VCVS gain). – G36 Dec 17 '20 at 21:15
  • And the second mesh equation should look like this: \$-V_\phi + 8\Omega I_2+4\Omega(I_2 - I_1)=0\$ – G36 Dec 17 '20 at 21:19
  • Yes, I think so, too. However the solution was different but as you said it might be a mistake, so thank you – JuanDoe Dec 17 '20 at 21:22
  • Also noticed that the control voltage (in blue) for your VCVS is equal to \$V_\phi = 4\Omega I_1\$ – G36 Dec 17 '20 at 21:26

1 Answers1

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Your first loop equation is correct. For the dependent source, only the i1 current needs to be taken into consideration. We can use the outer loop for finding Vth.

eqns

Kaswechiha
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